2
$\begingroup$

Question:

$\displaystyle\lim\limits_{x\to\pi/2}\frac{\ln(\sin x)}{\cos x}$

This was in a set of questions about l'Hôpital's rule.

However, the numerator is undefined and the denominator equals $0.5$ at $x$ rather than $0/0$ or $\infty/\infty$. How would I solve this?

$\endgroup$
  • 1
    $\begingroup$ $\cos(\pi/2)=0$. Why 0.5? $\endgroup$ – Przemysław Scherwentke Nov 20 '14 at 0:30
  • $\begingroup$ Oh what. I didn't have a calculator on hand and google told me it was .5. thanks! P.S. since the top is undefined does that mean I can treat it as 0? Since it approaches 0. $\endgroup$ – user3362196 Nov 20 '14 at 0:30
  • $\begingroup$ Likewise $\ln \sin (\pi/2) = \ln 1 = 0$. Also, Windows has a calculator, calc.exe. $\endgroup$ – Graham Kemp Nov 20 '14 at 0:54
  • $\begingroup$ @user3362196 Google told you it was $0.5$? Strange—I typed cos(pi/2) into Google just now, and it gave me $0$. (Perhaps you accidentally typed it cos(pi/3)?) $\endgroup$ – Akiva Weinberger Nov 20 '14 at 0:55
4
$\begingroup$

Both the numerator and the denominator are $0$, so it is an indeterminate form. If you are unsure about this, look at the sine and cosine waves.

You can apply L'Hopital's rule to the function:

$$\lim_{x\to \frac{\pi}{2}} \frac{\ln(\sin(x))}{\cos(x)}=\lim_{x\to\frac{\pi}{2}}\frac{\cos x}{\sin x \times -\sin x}=\lim_{x\to \frac{\pi}{2}}\frac{-\cos x}{\sin^2 x}$$.

Can you continue?

$\endgroup$
1
$\begingroup$

Yes,

$$ \lim_{x \rightarrow \frac{\pi}{2}}\frac{\ln(\sin(x))}{\cos(x)} = \lim_{x \rightarrow \frac{\pi}{2}}\frac{\cos(x)}{-\sin^2(x)} $$

Therefore

$$ \lim_{x \rightarrow \frac{\pi}{2}}-\cot(x) \csc(x) = -\cot(\frac{\pi}{2}) \csc(\frac{\pi}{2})= 0 $$

$\endgroup$
  • $\begingroup$ Hmm are you sure? My prof says that it should be 0. $\endgroup$ – user3362196 Nov 20 '14 at 0:33
  • $\begingroup$ I did it wrong at first, sorry for the confusion $\endgroup$ – Eric L Nov 20 '14 at 0:34
  • $\begingroup$ What's the point in using the cosecant and the cotangent? It's just unnecessary complication: $\cos(\pi/2)=0$ and $-\sin^2(\pi/2)=-1$. $\endgroup$ – egreg Nov 20 '14 at 0:38
  • $\begingroup$ @egreg What do you mean? $$ \frac{\cos(x)}{-\sin^2(x)} = -\cot(x) \csc(x) $$ $\endgroup$ – Eric L Nov 20 '14 at 0:38
  • $\begingroup$ I have a habit of expanding things out like so, its another perspective and it doesn't hurt. $\endgroup$ – Eric L Nov 20 '14 at 0:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.